Evaluate the following integrals: ^ -1 (4x^3-3x)dx

Question

Evaluate the following integrals:
$\int\cos^{-1}(4\text{x}^3-3\text{x})\text{dx}$

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Answer

Let $\text{I}=\int\cos^{-1}(4\text{x}^3-3\text{x})\text{dx}$
Let $\text{x}=\cos\theta$
$\text{dx }=-\sin\theta\text{d}\theta$
$\text{I}=-\int\cos^{-1}(4\cos^3\theta-3\cos\theta)\sin\theta\text{d}\theta$
$=-\int\cos^{-1}(\cos3\theta)\sin\theta\text{d}\theta$
$=-\int3\theta\sin\theta\text{d}\theta$
$=-3[\theta\int\sin\theta\text{d}\theta-\int(1\int\sin\theta\text{d}\theta)\text{d}\theta]$
$=-3[-\theta\cos\theta+\int\cos\theta\text{d}\theta]$
$=3\theta\cos\theta-3\sin\theta+\text{C}$
$\text{I}=3\text{x}\cos^{-1}\text{x}-3\sqrt{1-\text{x}^2}+\text{C}$

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